BTGmoderatorLU wrote:Source: Veritas Prep
A number when divided by 4 and 5 leaves remainders 1 and 4 respectively. What will be the remainder when this number is divided by 20?
A. 0
B. 3
C. 4
D. 9
E. 17
Now let´s avoid street fighting and embrace BJJ, a.k.a real MATH !
$$\eqalign{
& N = 4Q + 1 \cr
& N = 5K + 4 \cr} $$
$$? = R\,\,\,\left( {0 \le R \le 19} \right)\,\,\,,\,\,{\rm{where}}\,\,\,N = 20J + R\,\,\,\left( {J\,\,{\mathop{\rm int}} } \right)\,$$
$$Q,K,J,R\,\,\,{\rm{ints}}$$
$$\left\{ \matrix{
N = 4Q + 1\,\,\,\, \Rightarrow \,\,\,\,5N = 20Q + 5 \hfill \cr
N = 5K + 4\,\,\, \Rightarrow \,\,\,\,4N = 20K + 16 \hfill \cr} \right.\,\,\,\,\mathop \Rightarrow \limits^{\left( - \right)} \,\,\,\,\,\,\,N = 20\left( {Q - K} \right) - 11\,\,\,\,\,\, \Rightarrow \,\,\,\,N - 9 = 20\left( {Q - K} \right) - 20 = 20\left( {Q - K - 1} \right)\,$$
$$N - 9 = 20J\,\,,\,\,\,J = Q - K - 1\,\,{\mathop{\rm int}} \,\,\,\,\, \Rightarrow \,\,\,\,\,N = 20J + 9\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,? = 9\,\,$$
This solution follows the notations and rationale taught in the GMATH method.
Regards,
Fabio. 
